3 years ago

Round 73,588 ten thousand

2 answers:

Natasha2012 [34]3 years ago

6 0

70,000 because the 3 from 1 thousand is not higher than 5 so you keep it the same and change everything behind it to a zero

Alex73 [517]3 years ago

3 0

70,000 because 7 is in the ten thousand

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Order from least to greatest ! 1.36,1.3,1.63 and this other 0.42,3.74,3.47.

frosja888 [35]

If you are putting them all in order from least to greatest it would be 0.42, 1.3, 1.36, 1.63, 3.47,3.74. But, if you are separating them the first set of three numbers you wrote would be ordered 1.3, 1.36, 1.63. Then, the second set would be ordered 0.42, 3.47, 3.74. Hope this helped!!!

4 0

3 years ago

Read 2 more answers

Please help thanks!!

Lostsunrise [7]

Help with what there is nothing?

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3 years ago

what is the quotient 5-x/x^2 3x-4 divided by x^2-2x-15/x^2 5x 4 in simplifed form state any restrictions on the varible

zheka24 [161]

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

$\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$

$=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}$